Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2}{x - 1} = \dfrac{-3x + 4}{x - 1}$
Multiply both sides by $x - 1$ $ \dfrac{x^2}{x - 1} (x - 1) = \dfrac{-3x + 4}{x - 1} (x - 1)$ $ x^2 = -3x + 4$ Subtract $-3x + 4$ from both sides: $ x^2 - (-3x + 4) = -3x + 4 - (-3x + 4)$ $ x^2 + 3x - 4 = 0$ Factor the expression: $ (x - 1)(x + 4) = 0$ Therefore $x = 1$ or $x = -4$ However, the original expression is undefined when $x = 1$. Therefore, the only solution is $x = -4$.